FFT/NTT [51Nod 1028] 大数乘法 V2

题目链接：51Nod 传送门

没压位,效率会低一点

1.FFT

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int MAXN = 1<<18;
const double Pi = acos(-1.0);

struct complex
{
	double r, i;
	complex(double _r=0, double _i=0):r(_r), i(_i){}
	complex operator +(const complex &t)const
	{
		return complex(r + t.r, i + t.i);
	}
	complex operator -(const complex &t)const
	{
		return complex(r - t.r, i - t.i);
	}
	complex operator *(const complex &t)const
	{
		return complex(r*t.r - i*t.i, r*t.i + t.r*i);
	}
}a1[MAXN], a2[MAXN], w, wn;
char s1[MAXN], s2[MAXN];
int n, len1, len2, ans[MAXN];

inline void change(complex arr[], int len)
{
	for(int i = 1, j = len/2, k; i < len-1; ++i)
	{
		if(i < j) swap(arr[i], arr[j]);
		for(k = len/2; k <= j; j-=k, k>>=1);
		j += k;
	}
}

inline void FFT(complex arr[], int len, int flg)
{
    change(arr, len);
	for(int i = 2; i <= len; i<<=1)
	{
		wn = complex(cos(Pi*flg*2/i), sin(Pi*flg*2/i));
		for(int j = 0; j < len; j+=i)
		{
			w = complex(1, 0);
			for(int k = j; k < j + i/2; ++k)
			{
				complex wA1 = w * arr[k + i/2];
				complex A0 = arr[k];
				arr[k] = A0 + wA1;
				arr[k + i/2] = A0 - wA1;
				w = w * wn;
			}
		}
	}
	if(flg == -1)
		for(int i = 0; i < len; ++i)
			arr[i].r /= len;
}

int main()
{
	scanf("%s", s1), len1 = strlen(s1);
	scanf("%s", s2), len2 = strlen(s2);
	int len = len1 + len2;
	for(n = 1; n < len; n<<=1);

	for(int i = 0; i < len1; ++i) a1[i] = complex((double)(s1[i] - '0'), 0);
	for(int i = len1; i < n; ++i) a1[i] = complex();
	FFT(a1, n, 1);

	for(int i = 0; i < len2; ++i) a2[i] = complex((double)(s2[i] - '0'), 0);
	for(int i = len2; i < n; ++i) a2[i] = complex();
	FFT(a2, n, 1);

	for(int i = 0; i < n; ++i) a2[i] = a1[i] * a2[i];
	FFT(a2, n, -1);

	for(int i = 0; i < len1+len2-1; ++i)
		ans[i] = (int)(a2[i].r + 0.5);

	for(int i = len1+len2-2; i; --i)
	{
		ans[i-1] += ans[i]/10;
		ans[i] %= 10;
	}
	int i; for(i = 0; !ans[i] && i < len1+len2-1; ++i);

	if(i == len1+len2-1) putchar('0');
	else while(i < len1+len2-1) printf("%d",ans[i++]);
	putchar(10);
}

2.NTT

似乎并没有比FFT快

空间到是少了不少

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int MAXN = 1<<18;
const int mod = 998244353, G = 3;
char str1[MAXN], str2[MAXN];
int a1[MAXN], a2[MAXN];

inline void change(int arr[], const int& len)
{
	for(register int i = 1, j = len/2; i < len-1; ++i)
	{
		if(i < j) swap(arr[i], arr[j]);
		int k = len/2;
		while(j >= k) j -= k, k>>=1;
		j += k; //-.-写成了减样例半天没调过
	}
}

inline int qmul(int a, int b)
{
	int ret = 1;
	while(b)
	{
		if(b & 1) ret = (LL)ret * a % mod;
		a = (LL)a * a % mod; b>>=1;
	}
	return ret;
}

inline void NTT(int arr[], const int& len, const int& flg)
{
	change(arr, len);
	for(int i = 2, w, wn; i <= len; i<<=1)
	{
		if(~flg) wn = qmul(G, (mod-1)/i);
		else wn = qmul(G, mod-1 - (mod-1)/i);
		for(int j = 0; j < len; j += i)
		{
			w = 1;
			for(int k = j; k < j + i/2; ++k)
			{
				int A0 = arr[k];
				int wA1 = (LL)w * arr[k + i/2] % mod;
				arr[k] = (A0 + wA1) % mod;
				arr[k + i/2] = ((A0 - wA1) % mod + mod) % mod; //注意爆负

				w = (LL)w * wn % mod;
			}
		}
	}
	if(flg == -1)
	{
		int inv = qmul(len, mod-2);
		for(int i = 0; i < len; ++i)
			arr[i] = (LL)arr[i] * inv % mod;
	}
}

int main ()
{
	scanf("%s%s", str1, str2);
	int len1 = strlen(str1);
	int len2 = strlen(str2);
	int ml = len1 + len2, len = 1;
	while(len < ml) len<<=1;

	for(int i = 0; i < len1; ++i) a1[i] = str1[i] - '0';
	for(int i = len1; i < len; ++i) a1[i] = 0;
	for(int i = 0; i < len2; ++i) a2[i] = str2[i] - '0';
	for(int i = len2; i < len; ++i) a2[i] = 0;

	NTT(a1, len, 1), NTT(a2, len, 1);

	for(int i = 0; i < len; ++i) a2[i] = (LL)a1[i] * a2[i] % mod;

	NTT(a2, len, -1);

	for(int i = len1+len2-2; i; --i)
		a2[i-1] += a2[i] / 10, a2[i] %= 10;

	int i = 0;
	while(i < len1+len2-1 && !a2[i]) ++i;

	if(i == len1+len2-1) puts("0");
	else while(i < len1+len2-1) printf("%d", a2[i++]);
	putchar(10);
}